1. CSE 140 Lecture 8 Sequential Networks Professor CK Cheng CSE Dept. UC San Diego 1

2. Part II. Sequential Networks (Ch. 3) Memory / Time steps Clock Memory: Flip flops Specification: Finite State Machines Implementation: Excitation Tables xixi yiyi sisi y i =f i (S t,X) s i t+1 =g i (S t,X) 2

3. Memory Devices Memory Storage Latches Flip-Flops –SR, D, T, JK –State Tables –Characteristic Expressions 3

4. Memory Storage: Capacitive Loads Fundamental building block of other state elements Two outputs: Q, Q No inputs 4

5. Capacitive Loads Consider the two possible cases: –Q = 0: then Q’ = 1 and Q = 0 (consistent) –Q = 1: then Q’ = 0 and Q = 1 (consistent) –Bistable circuit stores 1 bit of state in the state variable, Q (or Q’ ) But there are no inputs to control the state 5

6. SR (Set/Reset) Latch SR Latch Consider the four possible cases: –S = 1, R = 0 –S = 0, R = 1 –S = 0, R = 0 –S = 1, R = 1 6

7. SR Latch Analysis –S = 1, R = 0: then Q = 1 and Q = 0 –S = 0, R = 1: then Q = 0 and Q = 1 7

8. SR Latch Analysis –S = 1, R = 0: then Q = 1 and Q = 0 –S = 0, R = 1: then Q = 0 and Q = 1 8

9. SR Latch Analysis –S = 0, R = 0: then Q = Q prev –S = 1, R = 1: then Q = 0 and Q = 0 9

10. S R y Q Q = (R+y)’ y = (S+Q)’ 10

11. Flip-flop Components S R SR F-F (Set-Reset) Inputs: S, R State: (Q, y) y Q 11

12. Id Q y S R Q* y Q** y** Q*** y*** 0 0 0 0 0 1 1 0 0 1 1 1 0 0 0 1 0 1 0 1 0 1 2 0 0 1 0 1 0 1 0 1 0 3 0 0 1 1 0 0 0 0 0 0 4 0 1 0 0 0 1 0 1 0 1 5 0 1 0 1 0 1 0 1 0 1 6 0 1 1 0 0 0 1 0 1 0 7 0 1 1 1 0 0 0 0 0 0 8 1 0 0 0 1 0 1 0 1 0 9 1 0 0 1 0 0 0 1 0 1 10 1 0 1 0 1 0 1 0 1 0 11 1 0 1 1 0 0 0 0 0 0 12 1 1 0 0 0 0 1 1 0 0 13 1 1 0 1 0 0 0 1 0 1 14 1 1 1 0 0 0 1 0 1 0 15 1 1 1 1 0 0 0 0 0 0 Q y State Transition SR 10 00 11 00 10 SR 11 10 01 11 01 11 01 1000 10 00 01 00 11 State Diagram 01 12

13. CASES: SR=01, (Q,y) = (0,1) SR=10, (Q,y) = (1,0) SR=11, (Q,y) = (0,0) SR = 00 => if (Q,y) = (0,0) or (1,1), the output keeps changing Solutions: 1) SR = (0,0), or 2) SR = (1,1). 0 0 0 1 - 1 1 0 1 - PS inputs 00 01 10 11 State table Q(t+1) SR Characteristic Expression Q(t+1) = S(t)+R’(t)Q(t) NS (next state) Q(t) 13

14. SR Latch Analysis –S = 0, R = 0: then Q = Q prev and Q = Q prev (memory!) –S = 1, R = 1: then Q = 0 and Q = 0 (invalid state: Q ≠ NOT Q) 14

15. SR Latch Symbol SR stands for Set/Reset Latch –Stores one bit of state (Q) Control what value is being stored with S, R inputs –Set: Make the output 1 (S = 1, R = 0, Q = 1) –Reset: Make the output 0 (S = 0, R = 1, Q = 0) Must do something to avoid invalid state (when S = R = 1) 15

16. D Latch Two inputs: CLK, D –CLK: controls when the output changes –D (the data input): controls what the output changes to Function –When CLK = 1, D passes through to Q (the latch is transparent) –When CLK = 0, Q holds its previous value (the latch is opaque) Avoids invalid case when Q ≠ NOT Q 16

17. D Latch Internal Circuit 17

18. D Latch Internal Circuit 18

19. D Flip-Flop Two inputs: CLK, D Function –The flip-flop “samples” D on the rising edge of CLK When CLK rises from 0 to 1, D passes through to Q Otherwise, Q holds its previous value –Q changes only on the rising edge of CLK A flip-flop is called an edge-triggered device because it is activated on the clock edge 19

20. D Flip-Flop Internal Circuit Two back-to-back latches (L1 and L2) controlled by complementary clocks When CLK = 0 –L1 is transparent, L2 is opaque –D passes through to N1 When CLK = 1 –L2 is transparent, L1 is opaque –N1 passes through to Q Thus, on the edge of the clock (when CLK rises from 0 1) –D passes through to Q 20

21. D Flip-Flop vs. D Latch 21

22. D Flip-Flop vs. D Latch 22

23. Latch and Flip-flops (two latches) Latch can be considered as a door CLK = 0, door is shutCLK = 1, door is unlocked A flip-flop is a two door entrance CLK = 1CLK = 0CLK = 1 23

24. D Flip-Flop (Delay) D CLK Q Q’ Id D Q(t) Q(t+1) 0 0 1 0 2 1 0 1 3 1 1 1 Characteristic Expression Q(t+1) = D(t) 0 0 1 1 0 1 PS D 0 1 State table NS= Q(t+1) 24

25. JK F-F J CLK Q Q’ 0 0 0 1 ? 1 1 0 1 ? PS JK 00 01 10 11 State table Q(t+1) K 25

26. JK F-F J CLK Q Q’ Characteristic Expression Q(t+1) = Q(t)K’(t)+Q’(t)J(t) 0 0 0 1 1 1 1 0 1 0 PS JK 00 01 10 11 State table Q(t+1) K 26

27. T CLK Q Q’ Characteristic Expression Q(t+1) = Q’(t)T(t) + Q(t)T’(t) 0 0 1 1 1 0 PS T 0 1 State table Q(t+1) T Flip-Flop (Toggle) 27

28. Using a JK F-F to implement a D and T F-F JKJK Q Q’ D CLK D flip flop JKJK Q Q’ T CLK T flip flop 28